Vol. 23, No. 2 (2017), pp. 33–45.
ON FULLY PRIME RADICALS
Indah Emilia Wijayanti
1and Dian Ariesta Yuwaningsih
21,2Department of Mathematics,
Universitas Gadjah Mada, Yogyakarta, Indonesia
1ind
−wijayanti@ugm.ac.id 2dian.ariesta17@yahoo.com
Abstract. In this paper we give a further study on fully prime submodules. For any fully prime submodules we define a product called∗M-product. The further
investigation of fully prime submodules in this work, i.e. the fully m-system and fully prime radicals, is related to this product. We show that the fully prime radical of any submodules can be characterized by the fully m-system. As a special case, the fully prime radical of a moduleMis the intersection of all minimal fully prime
submodules ofM.
Key words and Phrases: Fully invariant submodules, fully prime submodules, fully m-system, fully prime radicals.
Abstrak.Pada tulisan ini disajikan hasil penelitian terkait submodul prima penuh. Didefinisikan operasi perkalian-∗M untuk setiap submodul prima penuh. Penelitian
lebih lanjut terkait submodul prima penuh, yaitu sistem-mpenuh dan radikal prima
penuh, dapat dihubungkan dengan operasi perkalian-∗M ini. Lebih lanjut, dapat
ditunjukkan bahwa radikal prima penuh dari sebarang submodul dapat dibentuk dari sistem-mpenuh. Lebih lanjut, diperoleh bahwa radikal prima penuh dari suatu
modulMmerupakan irisan dari semua submodul prima penuh minimal diM.
Kata kunci: Submodul invarian penuh, submodul prima penuh, sistem-m penuh,
radikal prima penuh.
1. Introduction
Along with the process of generalization of rings to modules, some previous authors defined prime submodules as the generalization of prime ideals. By using some different approaches, there are some kinds of definition of prime submodules.
2000 Mathematics Subject Classification: 16D10.
34
The definition of prime submodules introduced by Dauns [2] has been referred by many authors for their study of primeness in module theory. Investigations of prime radicals and localization of modules by Wisbauer [11] are also based on this definition.
In this work we refer to the definition of primeness of submodules based on the paper of Wijayanti and Wisbauer [8], in which they defined prime (sub)module using the∗M-product and called it fully prime submodules. This∗M-product was
introduced earlier by Raggi et al. in their paper [7] and recently gives a possibility in a module to have such a ”multiplication”, such that we can adopt the definition of prime ideal to the module theory easier. The further study of the fully prime submodules started in paper Wijayanti [9]. The aim of this work is to complete the investigation of fully prime submodules, especially related to the m-system and radical.
Some previous authors also studied the prime submodules and prime radical submodules, for example Sanh [6], Lam [4] and Azizi [1]. We refer to Lam for defining fully m-systems and Sanh [6] for some properties of fully prime radicals of submodules.
In the next section we present some necessary and sufficient conditions of fully prime submodules in Proposition 2.2 and Proposition 2.5. Moreover, we show that there is a fully prime submodule which is not a prime submodule by giving a counter example in Example 2.3. To give some ideas of the fully primeness among another primeness, i.e. primeness in the sense of Dauns [2] and Sanh [5], we prove in Proposition 2.7, Proposition 2.9 and Lemma 2.10 that they not necessary coincide. Then we define a fully m-system and show that it is a complement of the fully prime submodule (see Proposition 2.13).
In the last section we define fully prime radicals and give some results. In Proposition 3.1 we characterize the fully prime radical of a submodule using the fully m-system. Moreover, in Proposition 3.9 we show that if the module is self-projective, then the fully prime radical of factor module modulo its fully prime radical is equal to zero.
ThroughoutRdenotes an associative ring with unit and the module should be a unital left module over the related ring. For our purpose, we write the homomor-phism on the right side. A fully invariant submoduleKinM is a submodule which satisfies (K)f ⊆K for any endomorphism f of M. Naturally, for anyR-module
M, it is also a rightS-module, where S= EndR(M) and the scalar multiplication
is defined asµ: (m, f)7→(m)f.
2. Fully Prime Submodules
We begin this section by the definition of a product between two fully invari-ant submodules as we refer to [7] and [8]. For any fully invariinvari-ant submodulesK, L
ofM, consider the product
We call the defined product in formulae (1) as the∗M-product. Now we recall the
definition of fully prime submodules in [8].
Definition 2.1. A fully invariant submoduleP of M is called a fully prime sub-module inM if for any fully invariant submodulesK, LofM,
K∗ML⊆P⇒K⊆P orL⊆P. (2)
A moduleM is calleda fully prime module if its zero submodule is a fully prime submodule.
Moreover, we give a characterization of fully prime submodules as we can show in the following proposition.
Proposition 2.2. Let P be a fully invariant submodule of M. The following as-sertions are equivalent:
a. P is a fully prime submodule;
b. For any m, k ∈ M and fully invariant cyclic submodules < m > and
< k >of M, if< m >∗M < k >⊆P, thenm∈P ork∈P.
c. For any submodules K, L of M, if KS∗M LS ⊆ P, then K ⊆ P or
L⊆P.
Proof. (a)⇒ (b). Letmand kbe elements in M where the fully invariant cyclic submodules < m > ∗M < k >⊆ P. Since P is fully prime, it implies < m >=
mS⊆P or< k >=kS⊆P. Thenm∈P or k∈P.
(b)⇒(c). LetK andLbe R-submodules ofM whereKS∗M LS ⊆P. Take any
k∈Kandl∈L, thenkS⊆KS andlS⊆LS. Moreover we have
kS∗M lS⊆KS∗MLS ⊆P,
and it impliesk∈P orl∈P. ThusK⊆P orL⊆P.
(c)⇒(a). LetK, Lbe fully invariant submodules ofM whereK∗ML⊆P. Since
KS ⊆K andLS ⊆L, thenKS∗M LS ⊆K∗M L⊆P. Thus we haveK ⊆P or
L⊆P.
Example 2.3. Let us give an example of a submodule which is not prime but fully prime. In Z-module Z12, we know that < 3 > is not a prime submodule. But
<3>is a fully prime submodule, since for any proper submodulesN, K inZ12, if
N∗Z12K⊆<3>, thenN ⊆<3>orK⊆<3>.
Now we recall some more properties according to Proposition 18 of Raggi et al. [7] which showed the relation between a fully prime submodule and the factor module. If a submoduleN is fully prime, then the factor moduleM/N is also fully prime. But we need some property for the converse, as we can see below.
Proposition 2.4. Let N be a proper fully-invariant submodule ofM.
(i) If N is fully prime inM, thenM/N is a fully prime module.
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Based on Proposition 2.4, if a moduleM is self-projective, then a submodule
P inM is fully prime if and only ifM/P is fully prime. LetLandU beR-modules and recall the following definition:
Rej(L, U) =\{Kerf |f ∈HomR(L, U)}
AnR-moduleLis called U-cogeneratedif Rej(L, U) = 0. For detailed explanation of reject and cogenerator, the readers are suggested to refer to Wisbauer’s book [10]. Moreover, we modify the result in 3.1 of [8] for a more general case.
Proposition 2.5. LetP be a fully invariant submodule of a self-projective module
M. The following statements are equivalent: a. P is a fully prime submodule;
b. Rej(M/P, K/P) = 0 for any non-zero fully invariant submodulesK/P
ofM/P;
c. K∗ML6⊆P for any non-zero fully invariant submodulesK/P andL/P
ofM/P;
d. Rej(−, M/P) = Rej(−, K/P) for any non-zero fully invariant submod-ulesK/P of M/P.
According to Proposition 2.5, ifP is fully prime, then anyM/P-cogenerated module is alsoK/P-cogenerated and vice versa.
ConsiderRas a leftR-module and letI, J be ideals ofR. ThenI∗RJ =IJ.
Since every ideal ofRis a fully invariantR-submodule, we get the following special case:
Proposition 2.6. [8]The following statements are equivalent for a two-sided ideal
I :
a. R/I is a prime ring.
b. I is a fully prime submodule in R. c. I is a prime ideal.
A module M satisfies the (∗f i) condition if for any non-zero fully invari-ant submodule K of M, AnnR(M/K) 6⊂ AnnR(M). A module M is called
fi-retractableif for any non-zero fully invariant submodule K ofM, HomR(M, K) =
AnnS(M/K)6= 0.
In general prime modules in the sense of Dauns [2] need not to be fully prime. Furthermore, in any self-projective module, if its submodule is prime, then it is not necessary fully prime. For the following relationship we generalize Proposition 3.4 of [8] as follows.
Proposition 2.7. For a self-projective R-moduleM with (∗f i)and for any fully invariant submodules K ofM, the following statements are equivalent :
Notice that for any ringR, EndR(R)≃Rand as a leftR-module,Rsatisfies
(∗f i) and is fi-retractable. IfM =R, Corollary 3.5 and Proposition 3.4 of [8] show that primeness and fully primeness ofRcoincide.
For any fully invariant submoduleKofM we denote (K:M) :={f ∈S |M f ⊆K}= AnnS(M/K).
Now we recall the definition of prime submodule in the sense of Sanh et al. [5], and we called it N-prime, as follows.
Definition 2.8. Let K be a fully invariant and proper submodule of M. The submoduleK is calledan N-prime submodules inM if for any fully invariant sub-modules L of M and for every ideal I of S, if I(L) ⊂ K, then I(M) ⊆ K or
L⊂K.
We refer to Theorem 1.2 of Sahn et. al. [5] to give a characterization of an N-prime submodule and consider that the prime notion in the sense of Sanh (N-prime) and endo-prime notion in the sense of Haghany-Vedadi [3] coinside. As an immediate consequence we extend Corollary 1.6 of [3] as follows.
Proposition 2.9. Suppose RM is fi-retractable and satisfies the (∗f i) condition.
ThenRM is prime if and only if RM is N-prime.
As a consequence of Proposition 2.7 and Proposition 2.9 we have the following property.
Lemma 2.10. Suppose RM is fi-retractable, satisfies the (∗f i) condition and is
self-projective. If M is fully prime, then M is N-prime.
Lemma 2.10 gives a sufficient condition of a fully prime module to be N-prime. Now we refer to Theorem 1.2 of Sahn et. al. [5] to prove the following property.
Proposition 2.11. LetM be anR-module, fi-retractable, satisfies the(∗f i) condi-tion and self-projective. Let S= EndR(M). If P is a fully prime submodule, then
(P :M)is a prime ideal in S.
Proof. Take anyf ∈S, an ideal T ofS where f T ⊆(P :M). Then (M)f T ⊆P. Moreover we have (M)f ST ⊆(M)f T ⊆P. According to Theorem 1.2 [5], since
P is fully prime, P is N-prime, i.e. (M)f S ⊆ P or M T ⊆ P. We show that
f S⊆(P :M) orT ⊆(P :M). Thus (P :M) is a prime ideal ofS.
Related to fully prime submodules, we observe now the notion below.
Definition 2.12. LetX be a non empty subset of a moduleM where 06∈X. X
is called afully m-systemif for anyx, y∈X,< x >∗M < y >∩X 6=∅.
As it is already known, there is a similar m-system notion in rings and it is a complement set of a prime ideal (see for example Lam [4]). In modules, we prove that the fully m-system is also a complement set of a fully prime submodule.
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Proof. (⇒). DenoteX =M\P. Take anyx, y∈X, thenx, y6∈P. By Proposition 2.2 (b),< x >∗M < y >6⊆P, hence< x >∗M < y >∩X 6=∅. Hence X=M \P
is a fully m-system.
(⇐). Take anyx, y6∈P, thenx, y∈M \P and< x >∗M < y >6⊆P. HenceP is
fully prime.
Next we show that any maximal fully invariant submodule is also a fully prime submodule.
Proposition 2.14. Let X be a fully m-system in M and P be a maximal fully invariant submodule in M whereP ∩X =∅. Then P is a fully prime submodule inM.
Proof. Suppose x6∈ P and y 6∈ P, but < x > ∗M < y >⊆ P. Then there exist
x1, x2∈X such thatx1∈P+< x >andx2∈P+< y >. Consider that
< x1>∗M < x2> ⊆ (P+< x >)∗M (P+< y >)
⊆ P+ (< x >∗M < y >)⊆P.
This is a contradiction with the fact thatX is a fully m-system.
Moreover, we also show that any fully prime submodule contains a minimal fully prime submodule.
Definition 2.15. A fully invariant submodule N in M is called minimal fully primeifN is minimal in the set of all fully prime submodules in M.
Proposition 2.16. Let P be a fully prime submodule in M. Then P contains a minimal fully prime submodule.
Proof. LetP be a fully prime submodule inM. We form the following set
J={U |U fully prime submodule inM whereU ⊆P}.
It is clear thatJ6=∅, sinceP ∈J. By Zorn’s Lemma,Jhas a minimal element or equivalently, every nonempty chain inJhas a lower bound inJ.
Consider a nonempty chainG⊆J. We construct a setQ= T
K∈G
K. It is clear that
Qis a fully invariant submodule in M andQ⊆P. We want to show thatQis a fully prime submodule inM. Take any two fully invariant submodulesX andY in
M whereX ∗M Y ⊆Qbut Y *Q. We prove thatX ⊆Q. Take any y∈Y \Q.
Then there exists K′
∈G such thaty 6∈K′
. SinceK′
is a fully prime submodule in M, fromX∗MY ⊆Q⊆K′ impliesX ⊆K′.
Then take anyL∈G. SinceGis a chain inJ,K′ ⊆
LorL⊆K′
. IfK′⊆
L, then
X ⊆ K′ ⊆
L. IfL ⊆K′
, then y 6∈ L. From X∗M Y ⊆Q ⊆L implies X ⊆L.
Thus X ⊆L for all L∈ G. Then X ⊆Q, and we prove thatQ is a fully prime submodule inM.
Since Q⊆P, Q ∈J and is a lower bound ofG. It is proved that any nonempty chain in J has a lower bound in J. Based on Zorn’s Lemma there exists a fully prime submodule P∗ ∈Jwhich is minimal inJ. Thus the fully prime submodule
P contains the minimal fully prime submoduleP∗.
LetN andKbe fully invariant submodules ofM whereK⊆N. We consider then the sets HomR(M, K) and HomR(N, K). For anyf ∈HomR(M, K), it induces
a homomorphism ˜f = f|N ∈ HomR(N, K). It is understood that there exists
an injective function from HomR(M, K) to HomR(N, K) which maps any f ∈
HomR(M, K) to ˜f =f|N ∈HomR(N, K). Hence, HomR(M, K)⊆HomR(N, K).
Proposition 2.17. Let N and P be a fully invariant submodules of M. If P is fully prime, then N∩P is a fully prime submodule in N.
Proof. Take any two fully invariant submodules K and Lin N, where L∗N K ⊆
N ∩P. Then L∗N K ⊆ P and LHomR(N, K) ⊆ P. Since HomR(M, K) ⊆
HomR(N, K),LHomR(M, K)⊆LHomR(N, K). MoreoverL∗M K⊆L∗NK and
henceL∗MK⊆P. ThenL⊆P orK⊆P becauseP is fully prime. ButK⊆N
andL⊆N, so we haveK⊆N∩P or L⊆N∩P as well.
We recall Lemma 17 of Raggi et al. [7] below.
Lemma 2.18. LetM andN beR-module andf ∈HomR(M, N)an epimorphism.
(i) If Kerf is a fully invariant submodule in M and L is a fully invariant sub-module inN, then(L)f−1 is a fully invariant submodule in M.
(ii) IfM is a quasi-projective module andU is a fully invariant submodule inM, then (U)f is a fully invariant submodule in N.
As a consequence, we have the following property.
Corollary 2.19. Let M be R-module and K,U submodules of M where K ⊂U
andKfully invariant inM. IfU/K is a proper fully invariant submodule inM/K, thenU is a proper fully invariant submodule inM.
Proposition 2.20. Let M be a self-projective module and A, P fully invariant submodules in M where A ⊂P. The submodule P is a fully prime submodule in
M if and only ifP/Ais a fully prime submodule in M/A.
Proof. LetS=EndR(M/A).
(⇒). We want to prove thatP/A is a fully prime submodule inM/A. Take any fully invariant submoduleK/AandL/AinM/A such that:
K/A∗M/AL/A=K/AHomR(M/A, L/A) =
X
f∈HomR(M/A,L/A)
(K/A)f ⊆P/A.
SinceK/AandL/Ais fully invariant submodules inM/A, according to Corollary 2.19 we get KandL are fully invariant inM.
Now take any f ∈ HomR(M/A, L/A) ⊆ S. Since M is a self-projective module,
there exists a homomorphismg∈HomR(M, L) such thatπf =gπ, whereπ is the
natural epimorphismπ:M →M/A. Then
(K/A)f =K(νf) =K(gν) = (Kg)ν = (Kg+A)/A⊆P/A.
Thus we haveKg⊆P.
40
Moreover, P
g∈S
Kg = KS ⊆ P. Since KHomR(M, L) = K∗M L ⊆ KS ⊆ P, it
implies K ⊆P or L ⊆P. We obtain K/A⊆P/Aor L/A ⊆P/A, i.e. P/Ais a fully prime submodule inM/A.
(⇐). Let P/A be a fully prime submodule in M/A. Take any fully invariant submoduleK andLin M where:
K∗M L⊆KHomR(M, L)⊆
X
f∈HomR(M,L)
Kf ⊆P.
Based on Lemma 2.18 (ii), we obtain that (K+A)/A and (L+A)/A are fully invariant submodules in M/A. Now take any f ∈ HomR(M, L) or equivalently
f ∈S. SinceAis a fully invariant submodule inM, we can construct the following function:
h:M/A → M/A
m+A 7→ (m+A)h=mf +A, for anym+A∈M/A.
Take any m+A, n+A∈M A andr∈R, then:
((m+A) + (n+A))h = ((m+n) +A)h
= (m+n)f +A
= (mf +nf) +A
= (mf +A) + (nf+A) = ((m+A)h) + ((n+A)h).
and
(r(m+A))h= (rm+A)h= (rm)f+A=r(mf +A) =r(m+A)h.
It is proved that his a homomorphism, so we geth∈S¯. Now take any m∈M. For any natural epimorphism ν:M →M Ait yields
m(νh) = (mν)h= (m+A)h=mf+A= (mf)ν =m(f ν).
It impliesνh=f ν.
Furthermore, P
f∈HomR(M,L)
Kf ⊆P impliesKf ⊆P. As a consequence,K(f ν) =
(Kf)ν ⊆P ν=P/A. We use the fact thatνh=f ν to obtainK(νh)⊆P/A. Then (K+A)A
h⊆P/Aand P
h∈S¯
(K+A)A
h= (K+A)A¯
S⊆P/A.
Since
(K+A)/AHomR(M/A,(L+A)/A)⊆((K+A)/A) ¯S⊆P/A,
we have ((K+A)/A)∗M/A((L+A)/A)⊆P/A. SinceP/Ais a fully prime submodule
in M/A, we get (K+A)/A ⊆ P/A or (K+A)/A ⊆ P/A. Hence K ⊆ P or
3. Fully Prime Radicals
Now we collect all fully prime submodules inM in the spectrum ofM below. We denote the fully prime radical of L as Radf pM(L), where L is a fully invariant submodule ofM.
Specf p(M) := {KM |K is fully prime submodule in M}
νf p(L) := {K∈Specf p(M)|L⊆K}
Radf pM(L) :=
\
K∈νf p(L)
K
In case a fully prime submodule which containsL does not exist, we define Radf pM(L) =M. It is easy to understand thatνf p(0) = Specf p(M) andνf p(M) =∅.
For a special case, we have Radf pM(0) =
T
K∈νf p(0)K. If M contains fully prime submodules, then
Radf pM(0) = \
K∈Specf pM(M)
K.
However, if there are no fully prime submodules in M, then Radf pM(0) =M. We
call the Radf pM(0) thefully prime radical ofM .
The following proposition gives us a characterization of the fully prime radical ofL.
Proposition 3.1. Let L be a fully invariant submodule ofM. Denote
R={x∈M | ∀fully m−systemX, x∈X ⇒X∩L6=∅}.
Then we have Radf pM(L) =M orRadf pM(L) =R.
Proof. First, we show that Radf pM(L)⊆ R. Take any y 6∈ R and prove that y 6∈ Radf pM(L). Since y 6∈ R, there is a fully m-system X′ such that y ∈ X′ but
X′∩L=∅. Consider now
M={N ⊆M |N is fully invariantL⊆N, N ∩X′=∅},
and a chainC inM. The set S
N∈MN is an upper bound of C. Hence by Zorn’s Lemma, there is a
maximal element in M, say N′. It is clear that N′ is a maximal fully invariant
submodule ofM andN′
∩X′
=∅. According to Proposition 2.14,N′
is fully prime andy6∈N′
. It impliesy6∈Radf pM(L).
Conversely, we show that R ⊆Radf pM(L). Take any x∈ RandN ∈νf p(L). Then
M \N is a fully m-system. Supposex∈M\N or equivalentlyx6∈N. It implies (M\N)∩L6=∅. ButL⊆N andN∩L6=∅, a contradiction. Thusx6∈M\N or
equivalentlyx∈N.
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Proposition 3.2. Let M be an R-module. Then Radf pM(0) = M or Radf pM(0) =
TP′
where theP′
s are minimal fully prime submodules of M.
Proof. If Radf pM(0)6=M, thenM contains a fully prime submodule and Specf p(M)6= ∅. According to Proposition 2.16, we know that any fully prime submodule inM
contains a minimal fully prime submodule. Then for any P ∈ Specf p(M), there exists a minimal fully prime submodule P′ ∈
Specf p(M) andP′ ⊆
P. Now we form the following set:
J={P′
∈Specf p(M)|P′
is a minimal fully prime submodule}.
We want to show Radf pM(0) =
Conversely, take any a6∈ Radf pM(0), then there exists a minimal fully prime sub-moduleP ∈Specf p(M) such that a6∈P. Consequentlya6∈ T
Proposition 3.3. Let{Lλ}be a family of fully invariant submodules inM. Then
\
that X is a fully prime submodule which contains Lλ for allλ∈Λ. HenceX also
contains the sumP
Proposition 3.4. Let M be a left R-modul, N and L fully invariant submodules of M. Thenνf p(N)∪νf p(L)⊆νf p(N∩L).
Proof. We have the following two cases. If Radfp
X ∈νf p(L), it means X a fully prime submodule inM where L⊆X. According
to Proposition 2.11 (X:M) is a prime ideal inS where (L:M)⊆(X:M). Then Radf pS ((L:M))⊆(X:M) and moreover
MRadf pS ((L:M))⊆M(X:M)⊆X.
We conclude then
MRadf pS ((L:M))⊆ \
X∈νf p(L)
X =Radfp
M(L).
Proposition 3.6. Let N be a fully invariant submodule of M. Then,Radfp
N(0)⊆
Radf pM(0).
Proof. Take any fully prime submodule P ∈ Specf p(M). If N ⊆ P, then the intersection of all fully prime submodule ofN is contained inP. Moreover we have Radf pN(0) ⊆ P. Now if N * P, then according to Proposition 2.17 we conclude
that N∩P is a fully prime submodule inN. Hence RadNf p(0)⊆N∩P ⊆P. It is
proved that Radf pN(0)⊆Radf pM(0).
Proposition 3.7. Let {Ni} be the family of fully invariant submodules of M. If
M =L
i∈I
Ni, a direct sum of fully invariant submodules Nis inM, then
M
i∈I
Radf pNi(0)⊆Rad
f p M(0).
Proof. It is obvious.
It is easy to prove the following property.
Proposition 3.8. Let M be an R-module andP1,P2 any fully prime submodules
inM. IfP1/Radf pM(0) andP2/Radf pM(0) are submodules inM/Rad f p
M(0), then:
P1/Radf pM(0)∩P2/RadMf p(0) = (P1∩P2)/Radf pM(0).
Proposition 3.9. If M is a self-projective module, then: Radf pM(M/Radf pM(M)) = ¯0.
Proof. IfM does not contain a fully prime submodule, then based on Proposition 2.20 we get thatM/Radf pM(0) also does not contain any fully prime submodule. So Radf pM =M and:
Radf pM(M/Radf pM(0)) = Radf pM(M/M) = Radf pM(¯0) = ¯0.
Now assumeM has a fully prime submodule. According to Proposition 2.20 we get that M/Radf pM(0) also contains a fully prime submodule. Then we have :
Radf pM(M/Radf pM(0)) = \
¯
P∈Specf p(M/Radf p M(0))
¯
44
We apply Proposition 3.8 to get the following:
Radf pM(M/Rad f p
M(0)) =
\
¯
P∈Specf p(M/Radf p M(0))
¯
P
= ( \
P∈Specf p(M)
P)Radf pM(0)
= Radf pM(0)/Radf pM(0) = ¯0
4. Concluding Remarks
Further work on the properties of fully prime radicals of submodules can be carried out. For example, we can define a fully multiplication module and then investigate the properties of fully prime radicals on the fully multiplication module. Moreover, we can bring this work to define fully prime localizations and then investigate the fully prime localization on fully multiplication modules.
Acknowledgement. We thank the referees for their time and comments.
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